Q.Find the mean deviation about mean for the following data: 4,7,8,9,10,12,13,17
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Mean Deviation About Mean
Mean Deviation About Mean – The Intuition First
Imagine you have a small set of numbers: the marks of five students in a test: 4, 6, 8, 10, 12. The average (mean) is 8. Now, each student is some distance away from this average. The student who scored 4 is 4 marks below the mean; the one who scored 12 is 4 marks above. The student who scored 8 is exactly at the mean.
If you simply add these distances, the positives and negatives cancel out — you get zero. That's not useful. So instead, we ask: on average, how far is each data point from the mean? That's the mean deviation about mean.
Mean deviation is a measure of spread or dispersion. It tells you how scattered the data is around the central value. A small mean deviation means most data points are close to the mean; a large one means they are spread out.
The Precise Definition
For a set of n observations x1,x2,…,xn with mean xˉ, the mean deviation about mean (often written as MD or M.D.) is:
MD(xˉ)=n1∑i=1n∣xi−xˉ∣
That vertical bars mean absolute value — we take the distance without caring about direction. So every deviation is positive.
Mean Deviation about Mean=n∑∣xi−xˉ∣
Step-by-Step Calculation
Let's use the marks example: 4, 6, 8, 10, 12.
Step 1: Find the mean.
xˉ=54+6+8+10+12=540=8
Step 2: Find each absolute deviation ∣xi−xˉ∣.
| xi | xi−xˉ | ∣xi−xˉ∣ |
|------|----------------|-------------------|
| 4 | -4 | 4 |
| 6 | -2 | 2 |
| 8 | 0 | 0 |
| 10 | 2 | 2 |
| 12 | 4 | 4 |
Step 3: Sum the absolute deviations.
4+2+0+2+4=12
Step 4: Divide by n=5.
MD=512=2.4
So, on average, each student's mark is 2.4 marks away from the mean of 8.
Notice that the mean deviation is always less than or equal to the standard deviation (another measure of spread). For this data, standard deviation is about 2.83, which is larger than 2.4. This is because standard deviation squares deviations, giving more weight to extreme values.
Why Use Absolute Values?
You might wonder: why not just average the plain deviations (without absolute value)? Because the sum of (xi−xˉ) is always zero — that's a property of the mean. The absolute value is the simplest way to make all deviations positive so they don't cancel.
A common mistake: forgetting to take absolute values and getting zero. Always check: if your sum of deviations is zero, you forgot the absolute value.
When Is This Used?
Mean deviation is intuitive and easy to explain. It's used in:
- Quality control (checking how consistent a manufacturing process is) …
First find the mean, then average the absolute deviations of each value from the mean. …
The mean deviation about the mean is 3.
Data: 4,7,8,9,10,12,13,17 (n=8).
Step 1 — Mean:
xˉ=84+7+8+9+10+12+13+17=880=10
Step 2 — Absolute deviations from the mean:
∣4−10∣=6, ∣7−10∣=3, ∣8−10∣=2, ∣9−10∣=1, ∣10−10∣=0, ∣12−10∣=2, ∣13−10∣=3, ∣17−10∣=7
Step 3 — Sum of deviations:
…
- CBSE 2025Set ANNUAL1 markMCQQ.Mean Deviation about mean of 5, 8, 9, 10, 11, 13, 14, 18 is:(a) 2.5(b) 3(c) 3.5(d) 4
›Reveal solutionSolution
Find the mean, take absolute deviations of each value from it, then average those deviations.
Data: 5,8,9,10,11,13,14,18 (n=8).
Mean xˉ=85+8+9+10+11+13+14+18=888=11
Absolute deviations ∣xi−xˉ∣: 6,3,2,1,0,2,3,7
…
- CBSE 2023Set ANNUAL1 markQ.Write the formula of mean deviation about mean for grouped data.
›Reveal solutionSolution
Mean deviation averages the absolute deviations of each class mark from the mean, weighted by frequency.
For grouped data with class marks xi, frequencies fi and N=∑fi, mean deviation about the mean xˉ is:
M.D.(xˉ)=N1∑ifi∣xi−xˉ∣ …
- CBSE 2023Set ANNUAL1 markQ.Fill in the blank: The formula for mean deviation from the mean, for ungrouped data, is ____.
›Reveal solutionSolution
For ungrouped data, mean deviation about the mean is n∑∣xi−xˉ∣.
Mean deviation measures the average absolute spread of the data from its mean.
…
- CBSE 2023Set ANNUAL1 markQ.What is the mean deviation from the mean of the data 3,4,5,6,7?
›Reveal solutionSolution
The mean deviation of 3,4,5,6,7 about their mean is 1.2.
Mean xˉ=53+4+5+6+7=525=5.
Absolute deviations: ∣3−5∣=2, ∣4−5∣=1, ∣5−5∣=0, ∣6−5∣=1, ∣7−5∣=2.
…
- CBSE 2022Set ANNUAL1 markQ.The mean deviation of 5, 6, 9, 11, 12, 3, 7, 11 about Mean is ............ .
›Reveal solutionSolution
The mean is 8; averaging ∣xi−8∣ over all 8 values gives a mean deviation of 2.75.
Data: 5,6,9,11,12,3,7,11 (n = 8)
Mean:
xˉ=85+6+9+11+12+3+7+11=864=8
Absolute deviations ∣xi−xˉ∣:
∣5−8∣=3, ∣6−8∣=2, ∣9−8∣=1, ∣11−8∣=3, ∣12−8∣=4, ∣3−8∣=5, ∣7−8∣=1, ∣11−8∣=3 …
- CBSE 2019Set ANNUAL1 markMCQQ.The mean deviation about mean of the following data 14, 17, 18, 19, 20, 22, 23, 27 is:(a) 20(b) 0(c) 3(d) √14
›Reveal solutionSolution
Compute the mean, then find n1∑∣xi−xˉ∣.
Data: 14,17,18,19,20,22,23,27 (n=8).
Mean xˉ=814+17+18+19+20+22+23+27=8160=20
Absolute deviations from mean: 6,3,2,1,0,2,3,7
…
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